Systems and Units

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Systems and Units Systems and Units The three systems of units are: 1. The English or the ft-lb-s System 2. The International or the m-kg-s System 3. The Laboratory or the cm-gm-s System Quantities fall into two main categories: 1. Principal Quantities Length (L) Mass (m) Time (t) Angle (θ) Temperature (T) 2. Derived Quantities Area (A) Volume (V) Velocity (ν) Frequency (Hz.) Acceleration (a) Angular Velocity (ω) Angular Acceleration (â) Volume Flow Rate (q) o Mass Flow Rate ( m ) Density (ρ) Specific Gravity (SG) Force (F) Force due to Inertia (FI) Force due to Gravity (FG) Force due to Viscosity (Fµ) Force due to Elasticity (FE) Force due to Pressure (FP) Specific Weight (γ) Energy (E) Moment of a Force (M) Work (W) Pressure (p) Stress (τ) Power (P) Dynamic Viscosity (µ) Kinematic Viscosity (ν) 1 1. Principal Quantities Quantity English System International System mile (mi) = 1760 yd kilometer (km) = 1000 m Length yard (yd) = 3 ft meter (m) = 100 cm foot (ft) = 12 in centimeter (cm) inch (in) = 2.54 cm Mass slug (sl) = 32.17404856 lbm kilogram (kg) = 1000 gm pound mass (lbm) = 453.592370 gm gram (gm) year = 365 d year = 365 d Time day = 24 h day = 24 h hour = 60 m hour = 60 m minute = 60 s minute = 60 s 1 rev. = 2B radians = 360E 1 rev. = 2B radians = 360E Angle 1 r = 57.29578E 1 r = 57.29578E 1 E = 0.017453r 1 E = 0.017453r Ordinary EF= 1.8EC+32 Ordinary EC= (EF-32)/1.8 Temperature Absolute ER= EF+459.688 Absolute K= EC+273.16 ER= 1.8 K K= ER/1.8 2 2. Derived Quantities Quantity Formula English Units SI Units Lab. Units mi2, ft2, in2, acre cm2, darcy, md Area A = L2 1 mi2 = 640 acres m2 1 cm2 = 1 acre = 43,560 101,320,790 d ft2 1 d = 1000 md bbl, ft3, gal, qt m3, ltr Volume V = L3 bbl = 5.6146 ft3 m3 = 1000 ltr cm3 = cc bbl = 42 gal ltr = 1000 cc gal = 4 qt = 3.785 ltr dL Velocity ν = ft/s m/s cm/s dt Frequency Hz. s-1 s-1 s-1 dν Acceleration a = ft/s2 m/s2 cm/s2 dt Angle 2 rad and E rad and E rad and E 360E = 2B rad 360E = 2B rad 360E = 2B rad dθ Angular ω = rad/s rad/s rad/s Velocity dt dω Angular aˆ = rad/s2 rad/s2 rad/s2 Acceleration dt Volume dV q = bbl/d, ft3/s m3/s, ltr/s cc/s Flow Rate dt o dm Mass Flow m = slug/s, lbm/s kg/s gm/s Rate dt m Density ρ = slug/ft3, lb/ft3, kg/m3 gm/cc V lb/gal Dw = 1.940 ρ 3 SG = f = slug/ft 3 Specific ρw Dw = 62.428 Dw= 10 Dw = 1 gm/cc 3 Gravity 141.5 lbm/ft3 kg/m 131.5+oAPI Dw = 8.345 lbm/gal Dw = 0.433 psi/ft Dw = 10 EAPI 3 2. Derived Quantities (cont.) Quantity Formula English Units SI Units Lab. Units Force F = m a lbf = slug ft/s2 N = kg m/s2 dyne = gm cm/s2 lbf = 444,822 dyne 1 N = 105 lbf = 4.44822 N dynes Force due to FI = m a lbf N dyne Inertia Force due to FG = m g lbf N dyne Gravity=Wt. g = 32.17404856 g = 9.80665 g = 980.665 ft/s2 m/s2 cm/s2 dν Force due to Fµ = µ A lbf N dyne Viscosity dy Force due to FE = E A lbf N dyne Elasticity Force due to FP = p A lbf N dyne Pressure 3 3 Specific Wt. ( = FG/V lbf/ft N/m dyne/cc Energy E = F L lbf-ft J = N-m erg = dyne cm Moment of a M = F L lbf-ft J erg = dyne cm Force Work W = F L lbf-ft J erg = dyne cm 2 2 2 2 Pressure p = FN/A lbf/ft , lbf/in = psi Pa = N/m dyne/cm 5 psc = 14.69594877 bar = 10 Pa psc=1013250 2 psi psc = 1.01325 d/cm bar psc = 1 atm psc = 101325 psc = 76 cm Hg Pa 2 2 2 Stress J = FT/A lbf/ft , lbf/in = psi bar, Pa dyne/cm Power P = E/t lbf-ft/s, hp W = J/s erg/s 1 hp = 550 lbf-ft/s F Dynamic µ lbf-s/ft2 = 47,880 N s/m2= 10 dyne s/cm2 = Fµ = Viscosity dν / dyA cp poise poise N s/m2= 1000 1 poise = 100 cp cp Kinematic < = µ/D ft2/s = 929 stoke m2/s = 104 cm2/s = stoke Viscosity stoke 4 EPS-441: Petroleum Development Geology Units and Conversion Semester: Homework #: Name: SS#: Problem #1: Do the following unit conversions: From To 38 oAPI lb/ft3 12 oAPI SG 56 oAPI psi/ft 40 oAPI SG 28 oAPI gm/cc 31 oAPI psi/ft 0.433 psi/ft lb/ft3 0.433 psi/ft lb/gal 0.433 psi/ft gm/cc 0.378 psi/ft oAPI 0.394 psi/ft oAPI SG = 0.76 psi/ft SG = 1.10 psi/ft SG = 0.74 lb/ft3 SG = 1.08 lb/gal SG = 0.88 oAPI 48.8 lb/ft3 SG 62.4 lb/ft3 oAPI 64.3 lb/ft3 psi/ft 48.7 lb/ft3 gm/cc 10.86 lb/gal gm/cc 8.33 lb/gal psi/ft 8.33 lb/gal lb/ft3 10.4 lb/gal gm/cc 0.82 gm/cc SG 1.02 gm/cc psi/ft 0.87 gm/cc psi/ft 0.91 gm/cc lb/ft3 Problem #2: Given a rectangular solid with dimensions 1000 ft x 400 ft x 40 ft. Calculate its volume in ft3, bbl, acre-ft?. 5 Problem #3: Calculate pressure gradients of the following liquids: SG = 1.00 3 Dw = 66.3 lb/ft SG = 1.15 Dw = 9.5 lb/gal SG = 0.85 EAPI = 42 EAPI = 32 EAPI = 45 3 Do = 58.1 lb/ft Problem #4: A well drilled to 3000 ft penetrates a formation containing 28 EAPI oil. If reservoir pressure is 1300 psia, what is the shut-in surface pressure?. If reservoir pressure is 1000 psia, how many ft of oil will be standing in the wellbore?. Problem #5: In an area where ambient temperature is 78 EF, two wells, A and B, were drilled. The depth of well A is 7250 ft and the depth of well B is 8000 ft. The bottom hole temperature (BHT) in well B is 180 EF. What is the BHT in well A?. Problem #6: The areal extent of a reservoir as determined with seismic data is 1500 acres. From logs the following reservoir properties were determined: Zone N h, ft Sw 1 0.28 4 0.28 2 0.32 7 0.40 3 0.18 3 0.31 4 0.20 11 0.27 a) Calculate the pore volume in the reservoir. Give your answers in acre-ft, ft3, and bbl. b) Calculate the volume of oil in the reservoir. Give your answers in acre-ft, ft3, and bbl. c) Calculate the STB of oil if Bo = 1.34 RB/STB. 6 Problem #7: The average reservoir properties of a 300 acre reservoir are: N = 18% and Sw = 36%. Estimated volume of oil originally in place = 410 MM STB. Formation thickness = 2000 ft. Determine the oil formation volume factor. Problem #8: Consider the sketch below. Given EAPI of oil = 35. water SG = 1.07. If well A penetrates the oil zone, find the shut-in surface pressure at well A?. 7 EPS-441: Petroleum Development Geology Units and Conversion Semester: Homework #: Name: SS#: Problem #1: A reservoir has an areal extent of 500 acres, an average thickness of 90 ft, and an average porosity of 20%. a) What is the reservoir volume available for hydrocarbons?. Answer in acre-ft, bbl, and ft3. b) If the average water saturation is 35%, what is the reservoir volume available for oil?. Answer in acre-ft, bbl, and ft3. c) Same as (b) except reservoir fluid is gas. d) If Bo = 1.34 RB/STB. What is the oil volume from (b) in surface barrels?. in surface cubic feet?. e) If Bg = 310 SCF/CF. What is the gas volume from (c) in surface cubic feet?. in surface barrels?. Problem #2: A well is being drilled to a depth of 10,000 ft. The wellbore diameter is 9 inches. The drill string has an inside diameter of 4.5 inches and an outer diameter of 5 inches. What is the volume of mud in the hole when it is at total depth (TD) and there is a complete drill string in the hole?. Ignore collars and bit volume and assume the mud is incompressible. Answer in cubic feet and barrels. 8 Problem #3: Data for the diagram shown below are: a) oil gravity = 38 EAPI b) water specific gravity = 1.05 c) average gas gradient = 0.055 psi/ft d) only oil flows in oil zone e) reservoir B is in water communication with the surface f) pressure at oil zone datum in reservoir A = 3820 psig g) temperature gradient = 1.5 EF/100 ft. h) mean surface temperature = 61 EF i) ground level is 3000 ft above sea level j) elevations given are sub-sea but answers should be in subsurface a) Calculate pressure and temperature at each of the 7 zones indicated in the diagram.
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